It was stated in this question (Give an intuitive explanation for polynomial quotient ring, or polynomial ring mod kernel) that given a field $F$, a general element of $F[x]/(a_n x^n + \ldots + a_0)$ is of the form $c_{n - 1}x^{n - 1} + \ldots + c_0$, with $x^n = \frac{a_{n-1}x^{n-1} + \ldots + a_0 x}{a_n}$. Now my question is, is this true if $F$ is not a field, but a commutative ring with unity/identity? And if so, how would one go about proving it?
Suppose that $R$ is a commutative ring with unity and let $f\in R[x]$ be a polynomial with degree $n \geq 1$. Then the ideal generated by $f$ is given by
$$ \langle f \rangle = \{gf| g\in R[x]\} $$
Then the quotient ring of this ideal is given by
$$ Q = R[x]/\langle f \rangle = \{[g] : g\in R[x]\} $$
where $[g]$ is syntactic sugar for $g + \langle f \rangle$. Now here I get stuck; how am I supposed to show that I can represent polynomials in $Q$ by polynomials with a degree that is less than $n$?
The reason I can't wrap my head around this is because the set $\langle f \rangle$ can contain polynomials with degree way larger than $n$ because we are taking every polynomial in $R[x]$ and multiplying it by $f$. Furthermore, in the definition of our cosets in the quotient ring, we are adding some polynomial from $R[x]$ to this product which may have degree greater than $n$, i.e if $[q]\in Q$ then $[q] = \{q + fg: g\in R[x]\}$ which again can give us polynomial of higher powers than $n$.
Also, the elements of $Q$ are cosets, not polynomials. How can we represent an element in $Q$ by a polynomial? I'm thoroughly confused.
Thanks for any help!
